Answer: at x = 0 the relative extrema is 1
Explanation:
The relative max and min occur when the derivative is zero.
Find the derivative of the function:
f(x) = x³ - 3x² + 1
f'(x) = 3x² - 6x + 0
Set the derivative equal to zero, factor, and solve for x:
3x² - 6x = 0
3x(x - 2) = 0
3x = 0 and x - 2 = 0
x = 0 and x = 2
One of these is the max (extrema) and the other is the min (minima). Plug these x-values into the original equation to find the y-values.
f(0) = (0)³ - 3(0)² + 1
= 0 - 0 + 1
= 1
f(2) = (2)³ - 3(2)² + 1
= 8 - 12 + 1
= -3
Since 1 is greater than -3, then 1 is the max and -3 is the min. (see attched graph).